Landau-Lifshitz pseudotensor - expressing the EM tensor ##T^{ik}##

[Kostik]

TL;DR

Landau-Lifshitz give an expression for the energy-momentum tensor $T^{ik}$ in terms of the metric and its first and second derivatives, and state that this is derivable "by simple transformations" from Einstein's equation $8\pi T^{ik} = R^{ik} - \frac{1}{2}g^{ik}R$. What are these "simple transformations"?

See the screen shot below from L-L "Classical Theory of Fields" 4th Ed. p. 281. L-L choose a point $x$, and work in locally inertial coordinates, so at the point $x$ the metric is constant: hence, $g_{ik,l}=0$. The EM tensor $T^{ik}$ can be written in terms of the metric (and its 2nd derivatives) via the Einstein equation $$8\pi T^{ik} = R^{ik} - \frac{1}{2}g^{ik}R$$ which appears in the middle of the page (setting $G=c=1$). In writing out the Ricci tensor $R_{ik}$, L-L discard the $\Gamma\Gamma$ terms because $g_{ik,l}=0$, and retain only the $\Gamma^a_{bc,d}$ terms. Thus, L-L give an expression for the contravariant Ricci tensor $R^{ik}$ in the middle of the page, and from this one can also express the Ricci scalar $R = g_{ik}R^{ik}$.

A little later, L-L remark, "After simple transformations the tensor $T^{ik}$ can be put in the form...", and you see the equation below.

Performing the two derivatives $\partial / \partial x^l$ and $\partial / \partial x^m$ on the expression shown will produce a horrendous mess. Likewise, writing out $R^{ik} - \frac{1}{2}g^{ik}R$ in terms of the metric will produce another horrendous mess.

What are the "simple transformations" that L-L is referring to?

 

[julian]

Applying certain identities, the expression for $R^{ik} - \frac{1}{2} g^{ik}R$ in terms of $g_{lp}$ (using the quoted expression for $R^{ik}$) just requires a few transformations to match the final form they provided. Below, I work backwards starting from their final expression.

First, some identities: It is fairly straightforward to show:

\begin{align*}
\frac{\partial}{\partial x^m} g = g g^{np} \frac{\partial g_{np}}{\partial x^m}
\end{align*}

Next, differentiating $g_{\alpha \mu} g^{\beta \mu} = \delta_\alpha^\beta$ implies

\begin{align*}
g_{\alpha \mu} \frac{\partial}{\partial x^m} g^{\beta \mu} = - g^{\beta \mu} \frac{\partial}{\partial x^m} g_{\alpha \mu}
\end{align*}

So that $g^{\alpha \beta}_{\;\; , m} = 0$ at the point $x$. Differentiating again gives:

\begin{align*}
g_{\alpha \mu} \frac{\partial^2}{\partial x^l \partial x^m} g^{\beta \mu} = - g^{\beta \mu} \frac{\partial^2}{\partial x^l \partial x^m} g_{\alpha \mu}
\end{align*}

at the point $x$. So that

\begin{align*}
\frac{\partial^2}{\partial x^l \partial x^m} g^{\alpha \beta} = - g^{\alpha n} g^{\beta p} \frac{\partial^2}{\partial x^l \partial x^m} g_{np}
\end{align*}

at the point $x$. Using these results:

\begin{align*}
& \frac{\partial}{\partial x^l} \frac{1}{(-g)} \frac{\partial}{\partial x^m} [(-g) (g^{ik} g^{lm} - g^{il} g^{km})]
\nonumber \\
& = \frac{\partial}{\partial x^l} [g^{np} g_{np,m}(g^{ik} g^{lm} - g^{il} g^{km}) + \frac{\partial}{\partial x^m} (g^{ik} g^{lm} - g^{il} g^{km})]
\nonumber \\
& = g^{np} g_{np,ml}(g^{ik} g^{lm} - g^{il} g^{km}) + \frac{\partial^2}{\partial x^l \partial x^m} (g^{ik} g^{lm} - g^{il} g^{km})
\nonumber \\
& = g^{np} g_{np,ml}(g^{ik} g^{lm} - g^{il} g^{km}) + g^{ik}_{\;\; ,ml} g^{lm} + g^{ik} g^{lm}_{\;\; ,ml} - g^{km} g^{il}_{\;\; ,ml} - g^{il} g^{km}_{\;\; ,ml}
\nonumber \\
& = g^{np} g_{np,ml}(g^{ik} g^{lm} - g^{il} g^{km}) - g^{lm} g^{i n} g^{k p} g_{n p , m l}
- g^{ik} g^{m n} g^{l p} g_{n p ,m l}
\nonumber \\
& + g^{km} g^{n i} g^{p l} g^{il}_{n p ,ml}
+ g^{il} g^{n m} g^{p k} g_{np ,ml}
\end{align*}

which is the same as $2R^{ik} - g^{ik} R$, as can be verified by comparing it with the quoted expression for $R^{ik}$:

\begin{align*}
2 R^{ik} & = g^{im} g^{kp} g^{ln} \left\{ g_{lp,mn} + g_{mn,lp} - g_{ln,mp} - g_{mp,ln} \right\}
\nonumber \\
& = (g^{im} g^{kp} g^{ln} + g^{ip} g^{km} g^{ln} - g^{im} g^{kl} g^{pn} - g^{in} g^{kp} g^{lm}) g_{np,ml}
\end{align*}

\begin{align*}
- g^{ik} R & = - \frac{1}{2} g^{ik} g^{mp} g^{ln} \left\{ g_{lp,mn} + g_{mn,lp} - g_{ln,mp} - g_{mp,ln} \right\}
\nonumber \\
& = g^{ik} g^{mp} g^{ln} \left\{ g_{ln,mp} - g_{lp,mn} \right\}
\nonumber \\
& = (g^{ik} g^{ml} g^{pn} - g^{ik} g^{mp} g^{ln}) g_{pn,ml}
\end{align*}

 

[Kostik]

Many thanks. Actually, I was able to prove it in the forward direction by writing the Ricci tensor and Ricci scalar in terms of the curvature tensor, thus: $$G^{\mu\nu}=R^{\mu\nu}-\frac{1}{2}g^{\mu\nu}R = \left( g^{\mu\alpha}g^{\nu\beta}g^{\rho\sigma}- \frac{1}{2} g^{\mu\nu}g^{\alpha\beta}g^{\rho\sigma} \right) R_{\rho\alpha\beta\sigma} \, .$$ Then there's a fair amount of work to do, but by careful grouping of terms, the Einstein equation $G^{\mu\nu}=-8\pi T^{\mu\nu}$ gives $$T^{\mu\nu} = \left[ (16\pi)^{-1}(g^{\mu\nu}g^{\alpha\beta}-g^{\mu\alpha}g^{\nu\beta})_{,\beta} \right]_{,\alpha}$$ which is the required form. (The expression in square brackets is $\nu$-$\alpha$ antisymmetric.)

 

[Kostik]

What now seems much more difficult is deriving Landau-Lifshitz's equations (96.8) and (96.9) from (96.7).

 

[julian]

What now seems much more difficult is deriving Landau-Lifshitz's equations (96.8) and (96.9) from (96.7).

Unfortunately, I don't have access to that part of the book. The Internet Archive version of L-L Volume 2 ends at page 183 for some reason.

 

[Kostik]

Unfortunately, I don't have access to that part of the book. The Internet Archive version of L-L Volume 2 ends at page 183 for some reason.

Try this:
https://archive.org/details/landau-l.-d.-lifshitz-e.-m.-course-of-theoretical-vol-2/page/n1/mode/2up

 

[julian]

What now seems much more difficult is deriving Landau-Lifshitz's equations (96.8) and (96.9) from (96.7).

Indeed — as they say, "after a rather lengthy calculation."